NCERT Solutions: Class 6 Maths Chapter 6 - Perimeter and Area

a. $2(l+2)=14$, so $l+2=7$ and $l=5\text{ cm}$. b. For a square, $4s=20$, so $s=5\text{ cm}$. c. $2(3+b)=12$, so $3+b=6$ and $b=3\text{ m}$.

The wire length is the rectangle perimeter: $2(5+3)=16\text{ cm}$. A square made from the same wire has side $16\div4=4\text{ cm}$.

Third side $=55-20-14=21\text{ cm}$.

Perimeter $=2(150+120)=540\text{ m}$. Cost $=540\times40=21600$, so the fencing costs $Rs\ 21,600$.

Divide the total string length by the number of equal sides: square $36\div4=9$, triangle $36\div3=12$, hexagon $36\div6=6$.

One round is the perimeter: $2(230+160)=780\text{ m}$. Three rounds require $3\times780=2340\text{ m}$.

Akshi's outer track perimeter is $2(70+40)=220\text{ m}$. In $5$ rounds she covers $5\times220=1100\text{ m}$.

Toshi's inner track perimeter is $2(60+30)=180\text{ m}$. In $7$ rounds she covers $7\times180=1260\text{ m}$, which is more than Akshi's $1100\text{ m}$.

A diagonal of a square is longer than its side. If diagonal boundary pieces are counted as one side unit, the perimeter is underestimated.

Because all sides of a regular polygon are equal, adding all side lengths is the same as multiplying one side length by the number of sides.

Width $=$ area $\div$ length $=300\div25=12\text{ m}$.

Area $=500\times200=100000\text{ sq m}$. There are $100000\div100=1000$ hundreds of square metres. Cost $=1000\times8=Rs\ 8000$.

Grove area $=100\times50=5000\text{ sq m}$. Number of trees $=5000\div25=200$.

Using Shape C as one unit of area, the areas are A = B = $4C$, C = E = $C$, and D = F = G = $2C$.

The hint states that Shape D can be covered exactly using Shapes C and E, and C and E have equal area. Thus $D=C+E=2C$.

Both Shape D and Shape F have area equal to $2$ times Shape C in the tangram.

Both Shape F and Shape G have area equal to $2$ times Shape C.

In terms of Shape C, $A=4C$ and $G=2C$, so $A=2G$.

Add all pieces in units of Shape C: $A+B+C+D+E+F+G=4+4+1+2+1+2+2=16$.

Rearranging the same seven pieces does not change their total area. Therefore the rectangle has the same area as the big square, $16C$.

The same pieces keep the same total area when rearranged, but the outside boundary can change. Therefore the perimeter of the square and the perimeter of a rectangle made from the same pieces need not be the same.

The total area is $5\times10+2\times7=50+14=64\text{ sq m}$. Any rectangle with area $64\text{ sq m}$ works.

Width $=$ area $\div$ length $=1000\div50=20\text{ m}$.

Floor area $=5\times4=20\text{ sq m}$. Carpet area $=3\times3=9\text{ sq m}$. Uncarpeted area $=20-9=11\text{ sq m}$.

Garden area $=15\times12=180\text{ sq m}$. Each flower bed has area $2\times1=2\text{ sq m}$, so four beds occupy $8\text{ sq m}$. Lawn area $=180-8=172\text{ sq m}$.

Shape A area $=2\times9=18$ and perimeter $=2(2+9)=22$. Shape B area $=4\times5=20$ and perimeter $=2(4+5)=18$. Thus Shape A has a smaller area but a longer perimeter.

If the page length is $L\text{ cm}$ and width is $W\text{ cm}$, the border rectangle has dimensions $(L-2)\text{ cm}$ by $(W-3)\text{ cm}$. Its perimeter is $2[(L-2)+(W-3)]=2L+2W-10\text{ cm}$.

The outer rectangle area is $12\times8=96$ square units. Half of this is $96\div2=48$ square units. Any inner rectangle not touching the boundary and having area $48$ square units satisfies the condition.

1. a. The area of each rectangle is larger than the area of the square.
2. b. The perimeter of the square is greater than the perimeters of both the rectangles added together.
3. c. The perimeters of both the rectangles added together is always 1 1/2 times the perimeter of the square.
4. d. The area of the square is always three times as large as the areas of both rectangles added together.

Let the square side be $s$. Each rectangle is $s\times\frac{s}{2}$, so each rectangle has perimeter $2(s+\frac{s}{2})=3s$. The two rectangle perimeters add to $6s$. The square perimeter is $4s$, and $6s=1\frac{1}{2}\times4s$.